318. Maximum Product of Word Lengths

Given a string array words, find the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0.

Example 1:
Given ["abcw", "baz", "foo", "bar", "xtfn", "abcdef"]
Return 16
The two words can be “abcw”, “xtfn”.

Example 2:
Given ["a", "ab", "abc", "d", "cd", "bcd", "abcd"]
Return 4
The two words can be “ab”, “cd”.

Example 3:
Given ["a", "aa", "aaa", "aaaa"]
Return 0
No such pair of words.

https://leetcode.com/discuss/74580/bit-shorter-c

判断两个字符串有没有共同的字符（都是小写字母）

class Solution {
public: 
    int maxProduct(vector<string>& words) {
        int result = 0;
        if(words.empty()) return result;
        unordered_map<int ,int> maps;
        for(auto word : words){
            int mask = 0;
            for(auto c : word)
                mask |= 1<<(c-'a');   //这句是精华，用比特位指示出现某个字符
            maps[mask] = max(maps[mask],(int) word.size());         
        }
        for(auto a : maps)
            for(auto b : maps){
                if(!(a.first & b.first))
                    result = max(result, a.second*b.second);
            }
        return result;
    }
};

转载于:https://www.cnblogs.com/CarryPotMan/p/5343672.html